To solve this problem, we need to analyze the group \( U_{37} \), which is the set of integers from \( 1 \) to \( 37 \) that are coprime to \( 37 \), under multiplication modulo \( 37 \). This is a standard group theory problem involving modular arithmetic.
The order of an element \(\bar{g}\) in a group is the smallest positive integer \( n \) such that \((\bar{g})^n \equiv 1 \pmod{37}\).
- First, we determine the elements of \( U_{37} \). Since \( 37 \) is a prime number, the elements of \( U_{37} \) are all integers from \( 1 \) to \( 36 \), because all these integers are coprime to \( 37 \).
- The group \( U_{37} \) has \( \varphi(37) = 36 \) elements, where \(\varphi\) is the Euler's Totient function.
- The possible orders of elements in \( U_{37} \) must divide the order of the group, which is \( 36 \). Hence, the possible orders are \( 1, 2, 3, 4, 6, 9, 12, 18, 36 \).
- Now, consider the statement: "The order of the element \( \overline{10} \) in \( U_{37} \) is 36." To check this, we compute powers of \( 10 \) modulo \( 37 \) to find the smallest positive integer \( n \) such that \( 10^n \equiv 1 \pmod{37} \).
- Let's compute small powers of \( 10 \) modulo \( 37 \):
- \( 10^1 \equiv 10 \pmod{37} \)
- \( 10^2 \equiv 100 \equiv 26 \pmod{37} \)
- \( 10^3 \equiv 260 \equiv 1 \pmod{37} \)
- Thus, the statement "The order of the element \( \overline{10} \) in \( U_{37} \) is 36." is indeed FALSE.
The other options would require separate validation, but based on the problem's context, the identified false statement aligns with our findings.